1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: LC Circuit inductance problem

  1. Dec 13, 2009 #1
    1. The problem statement, all variables and given/known data
    A circuit consists of a capacitor and an inductor. The resistance in the circuit is small and can be neglected. Initially, at t = 0 s, the voltage across the capacitor is at its maximum of VC = 10 V, the charge stored in the capacitor is 1 mC. It is observed that the capacitor discharges to QC = 0 C after t = 0.25 ms. What is the inductance L of the inductor in the circuit?

    http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys212/oldexams/exam3/fa09/fig8.gif [Broken]

    2. Relevant equations

    Conservation of energy:

    Energy Capacitor + Energy Inductor = Qmax^2 / 2C

    3. The attempt at a solution

    Initial Energy capacitor = .5*(Qmax^2/C) = .005
    Final Energy Capacitor = 0

    Initial Energy Inductor = 0
    Final Energy Inductor = .5*L*I^2


    .005 = .5*L*I^2 where I = dQ/dt which is (1E-3)/(.25E-3) = 4

    L = 6.25E-4 H which is wrong.

    Right answer is 253.3 microH.

    What am I doing wrong?
    Thanks for any help!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 13, 2009 #2


    User Avatar

    The inductor-capacitor circuit that you describe is analogous to a simple frictionless mass-spring system. It may be described by the following differential equation:

    [tex] L \frac{d^2q}{dt^2} + \frac{1}{C}q = 0 [/tex]

    The solution is [itex] q = q_m cos(\omega t + \phi) [/itex] which gives the charge on the capacitor at any time t.

    For the initial condition you describe, the constant [itex] \phi [/itex] is zero and

    [itex] \omega = \sqrt{\frac{1}{LC}} [/itex].

    Now, the capacitor discharges in .25 msec. What is the value of [itex] \omega t [/itex] when q = 0? (Hint: it has to make the cosine = 0.) That is sufficient for you to calculate L.

    Your mistake was assuming that the current is simply q/t.
    Last edited by a moderator: May 4, 2017
  4. Dec 13, 2009 #3
    Thank you thats perfect!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook